Question

A horizontal platform with an object placed on it is executing S.H.M. in the vertical direction. The amplitude of oscillation is 3.92x${10}^{-3}$. What must be the least period of these oscillations, so that the object is not detached from the platform

• A. 0.1256 sec
• B. 0.1356 sec
• C. 0.1456 sec
• D. 0.1556 sec

Answer 1

The correct option is A

0.1256 sec

By drawing free body diagram of subject during the downward motion at

extereme position, for equilibrium of mass

mg - R = mA ( A = Acceleration )

For critical condition R = 0

So mg = mA $\Rightarrow$ mg = m $\alpha {\omega }^2$

$\Rightarrow =\sqrt{\frac{g}{a}}=\sqrt{\frac{9.8}{3.92\times {10}^{-3}}}=50$